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Geometrical optics - lens floating in mercury

  1. Apr 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider a thin bi-convex lens with refractive index [itex]n[/itex] which has spherical surfaces with equal radii of curvature r and a measured focal length [itex]f[/itex]. The lens floats horizontally on the surface of liquid mercury so that its lower surface effectively becomes a spherical mirror. A point object on its optical axis is then found to coincide with its image when it is at a height [itex]u[/itex] above the lens. Determine [itex]r[/itex] and [itex]n[/itex].


    2. Relevant equations
    [tex]
    \frac{n_2}{v}+\frac{n_1}{u}=\frac{n_2-2_1}{R}
    [/tex]

    [tex]
    \frac{1}{u}+\frac{1}{v}=\frac{1}{f}
    [/tex]


    3. The attempt at a solution
    Here is my ray diagram for the situation:

    http://dl.dropbox.com/u/50942032/ray-diagram.JPG [Broken]

    The distance [itex]v[/itex] represents where the image would be formed if there was no lower surface to the lens (just refraction at a curved surface) so that the thin lens equation can be used. I've marked in some angles on the diagram:

    [itex]\alpha[/itex] represents the angle between the incident ray from the point object and the vertical.

    The other angles are angles which I've tried to use in formulating a solution - no success so far.

    I'm thinking that one approach would be to find the coordinates of intersection with the lower surface of the lens and then use geometry to find the coordinates of intersection with the top and then set up some simultaneous equations and solve them, but I haven't had any success with this approach.

    I'm guessing the small angle approximation can be used to approximate [itex]\sin\theta\approx\theta[/itex] in Snell's law and [itex]\tan\theta\approx\theta[\itex] in any geometry.

    The light will be refracted when it reaches the top surface of the lens (can apply Snell's law here), and then reflected on the bottom surface such that the angle of incidence equals the angle of reflection. Would it be possible to somehow use the refractive index of mercury (=1.000933) to work out the conditions for the critical angle - would this even help?

    I'm really stuck with this and would very much appreciate any help.

    With many thanks,

    Froskoy.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
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